Confusion on Variables - Excess of Variables

[AustrianSaurkraut]

Determine the value of a so that f(x)=√(ax2-4) has a tangent with a slope of 2 at the point where x equals to 2

For this, I started with the attempting the derivative, and got
f'(x) = 1/2(ax2-4)-1/2 * 2ax
From here, I began to sub in the X value but got stuck as there are too many A's

 

[topsquark]

First, some notation. The standard (text) formatting for raising something to a power is x^2, not x2. So your derivative would read
f'(x) = 1/2 * (ax^2 - 4)^(-1/2) * 2ax
(I use spaces alot, too.)

Okay. Let me write this in LaTeX for clarity. Canceling the 2 we get
[math] f'(x) = \dfrac{ax}{\sqrt{ax^2 - 4}}[/math]
Our requirement for f'(2) = 2 gives
[math]2 = \dfrac{2a}{\sqrt{4a - 4}}[/math]
You can do some canceling but here's the general idea. Isolate the square root, then square both sides:
[math]\sqrt{4a - 4} = \dfrac{2a}{2} = a[/math]
[math]4a - 4 = a^2[/math]
Now you have a quadratic in a.

-Dan